If the equation has two unequal real roots, then:

△=(k+ 1)^2-8(k+3)>； 0

k^2+2k+ 1-8k-24>； 0

k^2-6k-23>； 0

k^2-6k+9>； 32

(k-3)^2>； 32

K-3 > 4 √ 2 or k-3

K > 3+4 √ 2 or k < 3-√ 32

Let the two roots of the equation be a and b.

According to Vieta theorem

a+b=(k+ 1)/2

ab=(k+3)/2

According to the meaning, |a-b|= 1

So:

(a-b)^2= 1

(a+b)^2-4ab= 1

(k+ 1)^2/4-4*(k+3)/2= 1

(k+ 1)^2-8(k+3)=4

k^2+2k+ 1-8k-24-4=0

k^2-6k-27=0

(k+3)(k-9)=0

K=-3 or k=9

2. Substitute a=6-b into c2=ab-9 to obtain,

C2 = a b-9 =(6-b)b-9 = 6 b-B2-9 =-(b-3)2

∫C2≥0 and -(b-3)2≥0,

∴c=0 and b-3=0, which means c=0 and b=3.

∴a=6-b=6-3=3.